Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls . Find the probability that both the balls are red .
My solution goes like this:
We have $10$ black and $8$ red balls. The number of ways we can choose $2$ red balls is $\binom{8}{2}$ . The number of ways of choosing $2$ balls is $\binom{18}{2}$ . The required probability is $\frac{\binom{8}{2}}{\binom{18}{2}}$.
However if we do it in this way:
Considering two events $A$ and $B$ such that $A=$ the event of getting red ball in the 1st draw and $B=$ the event of getting red ball in the 2nd draw. Also, $P(A)$ and $P(B)$ are independent of each other. Now, $P(A)=\frac{8}{18}$ and $P(B)=\frac{8}{18 }$. So, $P(A\cap B)=P(A)P(B)=\frac{8^2}{18^2}=\frac{16}{81}$.
So which method is valid. Why is the other one not valid? Where is the problem occuring? I am not getting it.
Best Answer
The first solution corresponds to drawing the balls without replacement, which is not what you want.
For example, when you say that
you are not including the case where you draw the same ball twice.